Proof by working backwards(3)

The expression (x^4)/4 + (x + 1)^3 > ((x +1)^4)/4 for all real numbers x >= -1.

(Scratch work)

x^4 + 4(x^2 +2x +1)(x +1) > (x +1)^2(x^2 +2x +1)

x^4 + 4(x^3 +2x^2 +x +x^2 +2x +1) > (x+1)(x^3 +3x^2 +3x +1)

x^4 + 4(x^3 +3x^2 +3x +1) > x^4 +4x^3 +6x^2 +4x +1

x^4 + 4x^3 +12x^2 + 12x + 4 > x^4 + 4x^3 + 6x^2 + 4x + 1

6x^2 + 6x + 3 > 0

3x^2 +3x +1 > 0

Proof. Let f(x) = 3x^2 + 3x + 1, then f '(x) = 6x + 3 which implies that f attains its minimum value at x = -1/2. So for all x > -1,

f(x) >= 3/4 - 3/2 +1 = 1/4 > 0

Therefore we have that

3x^2 +3x +1 > 0 =>

6x^2 + 6x +3 > 0 =>

x^4 +  4x^3 +12x^2 + 12x +4 > x^4 +4x^3 +6x^2 +4x +1 =>

x^4 + 4(x^3 +3x^2 +3x +1) > x^4 +4x^3 +6x^2 +4x +1 =>

x^4 +4(x^3 +2x^2 + x + x^2 + 2x +1) > (x + 1)(x^3 + 3x^2 + 3x +1) =>

x^4 + 4(x^2 +2x + 1)(x + 1) > (x + 1)^2 (x^2 +2x + 1) =>

(x^4)/4 + (x + 1)^3 > ((x + 1)^4)/4

QED